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Q&A #9875


Forgiving Method/Partial Quotient Method of Division

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From: Gail (for Teacher2Teacher Service)
Date: Nov 12, 2002 at 23:42:35
Subject: Re: Forgiving Method/Partial Quotient Method of Division

Dear Pam,

Here is an example of an alternate algorithm that helps students who are
having difficulty with the traditional version.  It helps students gain an
understanding of what is happening during the division process, and assists
those who do not have a good recall of all the multiplication facts, because
they don't have to come up with the ONE correct digit for each step.
Instead, they can use a succession of factors to get to the final quotient.
Here is how it works...

   ____
7 /478     7 X 10     the student can select any factor they wish to use
  - 70                    to find a product to subtract from the original
  _____                       amount (the dividend)-- usually a student
   408     7 X 10                 selects an easy-to-use factor like
   -70                               1, 2, 5, or 10
  ____
   338     7 X 10
   -70                         As students become more familiar with the
  ____                              process they will select larger factors
   268     7 X 10
   -70
  ____
   198     7 X 10
   -70
  ____
   128     7 X 10
   -70
   ___
    58     7 X 2
   -14
   ___
    44     7 X 2
   -14
   ___
    30     7 X 2
   -14
  ____
    16     7 X 2    When you have found all the "groups" of 7 that can be
   -14              subtracted out, you are finished.  Now add all the
   ___              factors you multiplied by 7, and that will give you
     2                the quotient.

10 + 10 + 10 + 10 + 10 + 10 + 2 + 2 + 2 + 2 = 68

See, this student was able to finish the problem without using any of the
facts except 7 X 10 and 7 X 2.  This could be of great assistance to someone
who understood why we needed to divide, but didn't have the facts mastered
yet.

This is easier for struggling students because they don't have to have the
perfect factor each time, just one that will work.  They continue finding
factors and dividing until they have a remainder that is smaller than the
divisor (in this case, 7).  When they have found all the factors, they just
add them up (not the 7, but the other one each time).  That is their
quotient, and what is left is the remainder, which can be put in fraction
form (in this case, 2 out of 7, or 2/7) , or they can place a decimal, and
add zeroes to continue dividing.  Besides that help, this process treats the
dividend (in this case, 478) as a whole amount, instead of looking at 4
hundreds, then 48 tens, as the traditional algorithm does.  For some
students, that is a new way to look at the procedure.  They have been used
to looking at the amounts in terms of single digits...   "How many 7's are
in 4?"   then "How many sevens are in 48?"  No one has helped them see the
big picture, that the 4 is really 4 hundreds, and the 48, 48 tens...

I try to move my students to try larger "chunks" once we have that first
step mastered...
   ___
7 /478     7 X 30
  -210
  ____
   268     7 X 30
  -210
  ____
    58     7 X 5
   -35
  ____
    23     7 X 3
   -21
  ____
     2

30 + 30 + 5 + 3 = 68

Then we have a challenge to try to find the one "best chunk" for each place
value.  In the next example, there is one "chunk" for the tens, and one for
the ones.  (There is no hundreds chunk, because 7 X 100 would be more than
478).
   ___
7 /478     7 X 60
  -420
  ____
    58     7 X 8
   -56
  ____
     2

60 + 8 = 68

You will probably notice that that is VERY similar to the traditional
algorithm.  It is just a small nudge away for most students, at that point.
  :-)

 -Gail, for the T2T service

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